Pre-Quantum Electrodynamics

• Gr 3.3.1

Example: infinite grounded metal plates

Consider two infinite grounded metal plates positioned parallel to the \(xz\) plane, one at \(y = 0\) and the other at \(y = a\).

At \(x = 0\), the setup is closed off with an infinite insulated strip maintained at potential \(\phi_0 (y)\).

Question: find the potential inside the slot.

Solution:

By translational symmetry along \(z\), the potential must be independent of \(z\). This thus falls back onto a 2d problem. We need to solve the 2d Laplace equation Lap_2d

\[ \frac{\partial^2 \phi}{\partial^2 x} + \frac{\partial^2 \phi}{\partial^2 y} = 0 \]

\begin{align} &(i) ~\phi(x, y=0) = 0, \hspace{5mm} &(ii) ~\phi(x, y=a) = 0, \nonumber \\ &(iii) ~\phi(x=0, y) = \phi_0 (y), &(iv) ~\phi (x \rightarrow \infty, y) \rightarrow 0. \end{align}

Let us look for solutions in the form

\[ \phi(x,y) = X(x) Y(y), \hspace{1cm} \frac{1}{X} \frac{d^2 X}{dx^2} + \frac{1}{Y} \frac{d^2 Y}{dy^2} = 0 \tag{Lap_sv_car}\label{Lap_sv_car} \] Since the \(x\) and \(y\) dependencies are separated, the only possibility is that each individual term in the Laplace equation equals a constant, and that these constants add up to zero. We can thus put (the sign choice anticipates the solution somewhat)

\[ \frac{d^2 X_n}{dx^2} = k_n^2 X_n, \hspace{1cm} \frac{d^2Y_n}{dy^2} = -k_n^2 Y_n \tag{Lap_sv_car_sep}\label{Lap_sv_car_sep} \] where \(k_n\) is some real number. Since Lap_2d is a linear equation for \(\phi\), we can linearly combine solutions of Lap_sv_car_sep for different \(k_n\) and still get a solution to Lap_sv_car.

Let's look first of all at the solutions of Lap_sv_car_sep for a given \(k_n\). Since this is a second-order linear differential equation, there are two linearly independent solutions. The most general solution for \(X\) and \(Y\) can be written

\begin{align} X_n(x) &= Ae^{k_nx} + Be^{-k_nx}, \nonumber \\ Y_n(y) &= C \sin k_ny + D \cos k_ny \tag{Lap_sv_car_solXY}\label{Lap_sv_car_solXY} \end{align}

The constants can be fixed by fitting the boundary conditions. From \((iv)\), we get \(A = 0\). From \((i)\), we get \(D = 0\). We are thus left with

\[ \phi(x,y) = C_n e^{-k_nx} \sin k_n y \] Going further, \((ii)\) requires the momenta \(k_n\) to be quantized according to

\[ k_n = \frac{n\pi}{a}, \hspace{1cm} n = 1, 2, 3, ... \] Since we can use as solution any linear combination of the functions defined by these momenta, we obtain the solution in the form of a Fourier series,

\[ \phi(x,y) = \sum_{n=1}^{\infty} C_n e^{-n\pi x/a} \sin (n\pi y/a) \] with yet-to-be-determined coefficients \(C_n\), which have to be chosen to fit boundary condition \((iii)\). Using the orthogonality relation

\[ \int_0^a dy ~\sin(n\pi y/a) \sin (n' \pi y/a) = \delta_{n n'} \frac{a}{2} \] we thus get that the \(C_n\) are Fourier coefficients of the boundary function \(\phi_0(y)\):

\[ C_n = \frac{2}{a} \int_0^a dy ~\phi_0(y) \sin(n\pi y/a) \]

Specific example: say that \(\phi_0(y) = \phi_0\), i.e. just a constant. Then,

\[ C_n = \frac{2\phi_0}{a} \int_0^a dy \sin(n\pi y/a) = \frac{2\phi_0}{n\pi} (1 - \cos n\pi) = \frac{4\phi_0}{n\pi} \delta_{n, odd} \tag{Cn}\label{Cn} \]

Example: rectangular pipe

Take two infinitely long grounded plates at \(y = 0\) and \(y=a\), cut at \(x = \pm b\) and joined to metal strips maintained at constant \(\phi = \phi_0\).

Question: find the potential in the resulting rectangular pipe.

Solution: by translational invariance along \(z\), the potential must be independent of \(z\). We thus need to solve the 2d Laplace equation Lap_2d, \[ \partial^2 \phi/\partial x^2 + \partial^2 \phi/\partial y^2 = 0, \] with boundary conditions

\begin{align} &(i) ~\phi (x, y = 0) = 0, \hspace{5mm} &(ii) ~\phi (x, y = a) = 0, \nonumber \\ &(iii) ~\phi(x = b, y) = \phi_0, &(iv) ~\phi(x = -b, y) = \phi_0. \end{align}

The general solution is obtained from Lap_sv_car_XY, \[ \phi(x,y) = (Ae^{kx} + Be^{-kx}) (C\sin ky + D\cos ky). \] By symmetry, \(\phi(x,y) = \phi(-x,y)\) so \(A = B\). Since \(e^{kx} + e^{-kx} = 2\cosh kx\), the generic solution becomes (redefining \(C\) and \(D\)) \[ \phi(x,y) = \cosh kx ~(C\sin ky + D\cos ky). \] Boundary conditions \((i)\) and \((ii)\) require \(D = 0\), \(k = n\pi/a\) so

\[ \phi(x,y) = C \cosh(n\pi x/a) \sin(n\pi y/a) \] with \((iv)\) already satisfied if \((iii)\) is.

The full solution is then a linear combination of complete set of functions, \[ \phi(x,y) = \sum_{n=1}^{\infty} C_n \cosh(n\pi x/a) \sin(n\pi y/a). \] The coefficients must be chosen such that \((iii)\) is fulfilled, \(\phi(b,y) = \phi_0\). This simple case of a constant value \(\phi_0\) gives us the same relation as Cn, so

\[ \phi(x,y) = \frac{4\phi_0}{\pi} \sum_{n = 1, 3, 5, ...} \frac{1}{n} \frac{\cosh (n\pi x/a)}{\cosh(n\pi b/a)} \sin(n\pi y/a). \]